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VladimirAG [237]
3 years ago
10

If you had a total sample of 35 candium- 15 M&Mium, 12 Skittlitium and 8 MPeanuttium. What is the relative abundance for eac

h
Chemistry
1 answer:
Flura [38]3 years ago
4 0

<u>Given:</u>

Total Candium = 35

# of M&M = 15

# of Skittlitium = 12

# of Peanutttium = 8

<u>To determine:</u>

The relative abundance  for each

<u>Explanation:</u>

Relative abundance = amount of each item in the sample/Sample total

M&M = 15/35 = 0.429 (or) 42.9 %

Skittlitium = 12/35 = 0.343 (or) 43.3 %

Peanuttium = 8/35 = 0.229 (0r) 22.9%

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Describe where the "Reference Carbon" (the one that determines if the structure is D or L) is located in a carbohydrate with mor
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Farthest from the carbonyl carbon.

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5 0
3 years ago
At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react
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Answer:

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

N_{2 (g)} + O_{2 (g)} ⇄ 2NO_{(g)}

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no N_{2 (g)} nor  O_{2 (g)} present. Immediately, N_{2 (g)} andO_{2 (g)} are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [N_{2 (g)}]=0   ;     [O_{2 (g)} ]= 0    ; [NO_{(g)}]=0.60M

C: [N_{2 (g)}]=+x   ;     [O_{2 (g)} ]= +x    ; [NO_{(g)}]=-2x

E: [N_{2 (g)}]=0+x   ;     [O_{2 (g)} ]= 0+x   ; [NO_{(g)}]=0.60-2x

Now we can use the constant information:

K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }

4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}

4.10* 10^{-4}= \frac{(0.60-2x)^{2}}{x^{2} }

4.10* 10^{-4} * x^{2}= (0.60-2x)^{2}}

\sqrt{4.10* 10^{-4} * x^{2}}= \sqrt{(0.60-2x)^{2}}}

0.0202 x =0.60 - 2x

2x+0.0202x=0.60

x=\frac{0.60}{2.0202}= 0.30

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

3 0
3 years ago
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